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Homework Help: 3-dimensional potential energy problem

  1. Oct 25, 2004 #1
    I have a problem on my homework that says the potential energy of a particle is given by its position in the x-y plane according to

    P.E. = x^3 + 8x^2 + 34yz

    and I have to calculate the force on the particle at point (x,y,z), and all equilibrium points.

    dU = 3x^2 dx + 16x dx + 34y dz + 34z dy

    and F = -dU/dr with r being the vector position...

    -F = 3x^2 dx/dr + 16x dx/dr + 34y dz/dr + 34z dy/dr

    but that can't be the answer can it? I'm not given any dx/dr's or anything like that...I have a feeling there's some simple math thing I'm missing.

    Any hints anyone can think of?
  2. jcsd
  3. Oct 25, 2004 #2


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    You need to find the gradient of the potential - it's a vector quantity!
  4. Oct 25, 2004 #3
    Hmmm...I think my problem might be I've never learned how to take a vector derivative and this might be above my head mathematically :confused:

    Maybe I can separate the potential energy into x and y components...but i really don't know how.

    I guess since the particle is on the x-y plane, dz = 0...so my equation can be simplified to

    F = (-3x^2 - 16x) dx/dr - 34z dy/dr .

    Can I change dx/dr and dy/dr into (x-direction) and (y-direction) in terms of the force? Then use pythagorean theorem to find the total force? Am I just rambling?
  5. Oct 25, 2004 #4


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    It's really straightforward:

    The x-component of the force is the partial derivative of the potential with respect to x. For the y-component use the partial wrt y and for the z-component use the partial wrt z. Voila!
  6. Oct 25, 2004 #5


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    One caveat: the force is MINUS the gradient of the potential, [itex] \vec F = - \vec \nabla V [/itex]

    or, to be more specific,

    [itex] F_x = - {\partial V(x,y,z) \over \partial x} [/itex]

    [itex] F_y = - {\partial V(x,y,z) \over \partial y} [/itex]

    [itex] F_z = - {\partial V(x,y,z) \over \partial z} [/itex]

    If the particle is indeed constrained to the xy plane, then the z component of the force will be canceled by a normal force. In the x and y component, one must then set the value of z corresponding to the z plane one is in (z=0 or some other value).

  7. Oct 25, 2004 #6


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    Thanks - I meant to type "proportional to" but somehow it didn't come out!
  8. Oct 26, 2004 #7
    If I take the derivative with respect to, say, x, can I assume y and z are constants? If so, I messed up on a pretty simple problem :blushing:
  9. Oct 26, 2004 #8


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    That's essentially what you do when you take partial derivatives which applies here.
  10. Oct 27, 2004 #9
    Thanks for the help guys!
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